Thinkphp6 上传文件后如何获取新文件名

Question

<?php

namespace app\index\controller;
use think\facade\View;
use app\model\Exam as ExamModel;

class Exam extends Common
{
    public function exam()
    {
        return View::fetch();
    }
    public function uploadfile()
    {
        $file = request()->file('file');
        try {
            validate(['imgFile' => [
                'fileSize' => 410241024,
                'fileExt' => 'jpg,jpeg,pptx,zip',
                'fileMime' => 'image/jpeg',
            ]])->check(['imgFile' => $file]);
            $saveName = \think\facade\Filesystem::disk('public')->putFile('upload', $file, 'data');
//$saveName 新文件名 如何拿到下面的 upload方法中赋值给$data
            return ['code' => 0, 'msg' => '文件上传成功'];
        } catch (\Exception $e) {
            return ['code' => 1, 'msg' => $e->getMessage()];
        }
    }
    public function upload()
    {
        $data = request()->param();
        $data['uid'] = session('id');
        $data['status'] = 0;
        // $data['file'] = $saveName 在这里使用
        return (new ExamModel())->add($data);
    }
}

我在定义过一个类属性$arr，然后将$saveName 的值赋给$arr类属性，但是在upload中调用这个属性的$this->$arr 就变成nul应该如何操作呢。求指点！万分感谢！

Answer 1

存取session